Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx

... second equation. c 1 r 2 1 + 1 r 2 (1 − c 1 r 1 )r 2 2 = 1 c 1 (r 2 1 − r 1 r 2 ) = 1 − r 2 11 81 0.2 0 .4 0.6 0.8 1 1.2 0.3 0 .4 0.5 0.6 0.7 0.8 0.9 Figure 23.2: Plot of the solution and approximations. Recall ... that a n =   n/2 j=0  − 4j 2 −2j +1 (2j+2)(2j +1)  for even n, 0 for odd n. 11 93 c 1 = 1 − r 2 r 2 1 − r 1 r 2 = 1 − 1 √ 5 2...

Ngày tải lên: 06/08/2014, 01:21

... exponential and the logarithm.  √ x + 1 exp  ı √ 4 − 1 2 ln(x + 1)  , √ x + 1 exp  −ı √ 4 − 1 2 ln(x + 1)  . Note that  √ x + 1 exp  ı √ 4 − 1 2 ln  x + 1 2  , √ x + 1 exp  −ı √ 4 − 1 2 ln  x ... series. 1 4 = 1 6 − 1 π 2 ∞  n =1 cos(nπ) n 2 ∞  n =1 ( 1) n n 2 = − π 2 12 (c) We substitute x = 1/ 2 into the sine series. 1 4 = 8 π 3 ∞...

Ngày tải lên: 06/08/2014, 01:21

... + ı a =  1 s 4 − 1  s=0 = 1 b =  d ds 1 s 4 − 1  s=0 = 0 c =  1 s 2 (s + 1) (s − ı)(s + ı)  s =1 = 1 4 d =  1 s 2 (s − 1) (s − ı)(s + ı)  s= 1 = − 1 4 e =  1 s 2 (s − 1) (s + 1) (s + ı)  s=ı = ... + 1) . We solve for ˆy 1 . ˆy 1 = 6 s 4 (s 3 + s 2 + s + 1) ˆy 1 = 1 s 4 − 1 s 3 + 1 2(s + 1) + 1 − s 2(s 2 + 1) We then take t...

Ngày tải lên: 06/08/2014, 01:21

... imaginary part of some analytic function. Solution 8 .11 We write the real and imaginary parts of f(z) = u + ıv. u =  x 4/ 3 y 5/3 x 2 +y 2 for z = 0, 0 for z = 0. , v =  x 5/3 y 4/ 3 x 2 +y 2 for ... 8 .12 Consider the complex function f(z) = u + ıv =  x 3 (1+ ı)−y 3 (1 ı) x 2 +y 2 for z = 0, 0 for z = 0. Show that the partial derivatives of u and v with respect to...

Ngày tải lên: 06/08/2014, 01:21

... doubles every hour. For the continuous problem, we assume that this i s true for y(t). We write this as an equation: y  (t) = αy(t). 775 1 2 3 4 4 8 12 16 1 2 3 4 4 8 12 16 Figure 14 . 1: The p opulation ... = y(x) x . P (1, u) + Q (1, u)  u + x du dx  = 0 This equation is separable. P (1, u) + uQ (1, u) + xQ (1, u) du dx = 0 1 x + Q (1, u) P (1, u) + uQ (1, u) du dx =...

Ngày tải lên: 06/08/2014, 01:21

... + 1) for n = m.  1 1 P 0 (x)P 0 (x) dx =  1 1 1 dx = 2  1 1 P 1 (x)P 1 (x) dx =  1 1 x 2 dx =  x 3 3  1 1 = 2 3  1 1 P 2 (x)P 2 (x) dx =  1 1 1 4  9x 4 − 6x 2 + 1  dx =  1 4  9x 5 5 − ... 1) a n +1 z n 1 + 1 2z ∞  n=0 (n + 1) a n +1 z n + 1 z ∞  n=0 a n z n = 0 ∞  n =1  n(n + 1) a n +1 + 1 2 (n + 1) a n +1 + a n  z n 1 +...

Ngày tải lên: 06/08/2014, 01:21

... continuous. -1 -0.5 0.5 1 -1 -0.5 0.5 1 -1 -0.5 0.5 1 -1 -0.5 0.5 1 Figure 25 .1: Polynomial Approximations to cos(πx). 12 86 0 1 2 3 4 5 0.25 0.5 0.75 1 1.25 1. 5 1. 75 2 Figure 24. 1: Plot of K 0 (x) and ... (t  ) 2 − 2x 1/ 2 t  − 1 2 x 1/ 2 = 0 1 4 x −2 + u  +  − 1 4 x 1 + u   2 − 2x 1/ 2  − 1 4 x 1 + u   − 1 2 x 1/ 2 = 0 u  + (u...

Ngày tải lên: 06/08/2014, 01:21

... + ∞  n=−∞ n=0  1 in e ınt  x −π    = 1 2π    (x + π) + ∞  n=−∞ n=0 1 ın  e ınx −( 1) n     = x 2π + 1 2 + 1 2π ∞  n =1 1 ın  e ınx − e −ınx −( 1) n + ( 1) n  = x 2π + 1 2 + 1 π ∞  n =1 1 n sin(nx) 13 01 Alternatively, ... ( 1) n ın √ 2π 1 √ 2π e ınx = 1 2 + 1 π ∞  n =1 1 − ( 1) n n sin(nx) H(x) ∼ 1 2 + 2 π ∞  n =1 oddn 1 n sin(nx). Int...

Ngày tải lên: 06/08/2014, 01:21

... yields f  (x) ∼ 4 ∞  n =1 oddn cos(nx). For x = nπ, this implies 0 = 4 ∞  n =1 oddn cos(nx), 13 61 -1 -0.5 0.5 1 -0.2 -0 .1 0 .1 0.2 1 0.25 0 .1 0 0 .1 1 1 1 0.5 Figure 28.8: Three Term Approximation for a ... expansion. b n =  1/ 2 1 (−x − 1) sin(nπx) dx +  1/ 2 1/ 2 x sin(nπx) dx +  1 1/2 (−x + 1) sin(nπx) dx = 2  1/ 2 0 x sin(nπx) dx + 2  1 1/2...

Ngày tải lên: 06/08/2014, 01:21

... = a 0 2 n  k =1 a k cos(kx), where a k = 1 + ( 1) n−k 2 1 2 n 1  n (n − k)/2  . 13 88 3. A n + ıB n = n  k =1 z k = 1 − z n +1 1 − z = 1 − r n +1 e ı(n +1) x 1 − r e ıx = 1 − r e −ıx −r n +1 e ı(n +1) x +r n+2 e ınx 1 ... n 1 n 2 cos(nx)        ≤ 4 π ∞  n=N +1 odd n 1 n 2 = 4 π ∞  n =1 odd n 1 n 2 − 4 π N  n =1 odd n 1 n 2 Since ∞  n =1 o...

Ngày tải lên: 06/08/2014, 01:21